Free Steel Tension Member Calculator -- AISC 360

Design steel tension members including W-shapes, WT sections, double angles, single angles, HSS rounds and squares, threaded rods, and flat plates. The calculator checks all applicable tension limit states: yielding on gross cross-sectional area (AISC 360 D2), rupture on net effective area (D3), block shear at bolted end connections (D5 referencing J4.3), slenderness limits for serviceability (D1), and shear lag effects for members connected by only some cross-sectional elements (D3.3). Coverage spans AISC 360-22 Chapter D, AS 4100 Section 7, EN 1993-1-1 Section 6.2.3, and CSA S16 Section 13.

Tension member design is conceptually the simplest structural check -- the member is pulled axially and the cross-section resists -- but the interaction of connection geometry, bolt hole layout, and shear lag makes it one of the most frequently miscalculated limit states in practice. The critical insight is that a tension member is only as strong as its weakest net section, which occurs where bolt holes reduce the cross-sectional area to a local minimum. The effective net area further reduces this by the shear lag factor U, reflecting the non-uniform stress distribution at end connections.

Member types supported:

What this calculator does not cover: tension members with pin-connected ends (eye bars, pin plates -- see AISC 360 D5), tension members subject to fatigue loading (AISC 360 Appendix 3), tension rods with upset ends, and tension members with combined axial tension and flexure (beam-columns -- see AISC 360 Chapter H).

How to Use This Calculator

Step 1 -- Select member type and section. Choose the member type and enter the section designation (e.g., W10x45, 2L5x5x1/2, HSS6x6x1/2). The calculator retrieves dimensions and properties from the built-in shape database: gross area Ag, flange/web thicknesses, centroid location (x_bar and y_bar for angles and tees), and radius of gyration r_min for slenderness checks.

Step 2 -- Designate connection configuration. For bolted connections, specify the bolt diameter (d_b), number of bolts, bolt layout (number of lines, stagger pattern if staggered), and bolt hole type (standard, oversized, short-slotted, or long-slotted). The net area An = Ag - Sum(d_hole x t) for each element with bolt holes, where d_hole = d_b + 1/16 in for standard holes + 1/16 in for damage allowance = d_b + 1/8 in total. For welded connections, shear lag is typically minimal (U ≈ 1.0 for welds along the full member perimeter, U = 1 - x_bar/L for longitudinal welds on one side only).

Step 3 -- Specify steel grade and design method. Enter Fy and Fu (or select standard grade: A36 at 36/58 ksi, A572 Gr 50 at 50/65 ksi, A992 at 50/65 ksi). Select LRFD (phi = 0.90 yielding, 0.75 rupture, 0.75 block shear) or ASD (Omega = 1.67 yielding, 2.00 rupture, 2.00 block shear). The yielding-to-rupture resistance factor ratio (0.90/0.75 = 1.20) reflects the philosophy that yielding (a ductile, serviceable limit state) is penalized less than rupture (a brittle, catastrophic limit state).

Step 4 -- Apply factored tension demand. Enter the factored axial tension Pu (LRFD) or service tension Pa (ASD). The calculator compares Pu to phi_Pn for each limit state, reporting the governing demand-to-capacity ratio (DCR). For LRFD, DCR = Pu / (phi x Rn); for ASD, DCR = Pa / (Rn/Omega).

Step 5 -- Review all limit states. The output displays: (a) yielding DCR, (b) rupture DCR (several paths if staggered bolts create multiple potential rupture lines), (c) block shear DCR (all possible block shear paths -- tearing out through the bolt group and tension rupture at the end), and (d) slenderness ratio L/r compared to the 300 limit for tension members subject to stress reversal.

Step 6 -- Interpret the governing DCR. If rupture governs (DCR_rupture > DCR_yielding), the member has adequate gross area but the bolt hole pattern reduces the net section excessively. Solutions: increase the bolt stagger (staggered holes increase the net width via the s^2/(4g) rule), reduce the bolt diameter, or increase the member size. If yielding governs, the member needs a larger gross area. If block shear governs, increasing the end distance or bolt spacing typically resolves the issue.

Engineering Theory -- Tension Member Design

Gross Section Yielding (AISC 360 D2)

Yielding on the gross cross-section represents the onset of significant (unbounded) elongation of the member. The design strength for yielding is:

phi_Pn = phi x Fy x Ag    (LRFD: phi = 0.90)
Pn/Omega = Fy x Ag / 1.67  (ASD: Omega = 1.67)

Yielding is a ductile limit state -- the member elongates visibly before reaching its ultimate capacity, providing warning of distress. The 0.90 resistance factor reflects this reliability. Yielding typically governs for short, stocky members with few or no bolt holes.

Net Section Rupture (AISC 360 D3)

Rupture on the net effective area is a brittle limit state -- the member fractures suddenly with little warning at the bolt hole locations. The design strength for rupture is:

phi_Pn = phi x Fu x Ae    (LRFD: phi = 0.75)
Pn/Omega = Fu x Ae / 2.00  (ASD: Omega = 2.00)

where Ae = An x U is the effective net area. An is the net area deducting bolt holes from the gross area. For a row of bolts perpendicular to the member axis: An = Ag - n_holes x d_hole x t. For staggered bolts, the net width includes the s^2/(4g) term per AISC 360 Section B4.3b:

wn = wg - Sum(d_hole) + Sum(s^2/(4g))

where s is the longitudinal stagger (pitch) and g is the transverse spacing (gage). The s^2/(4g) term accounts for the fact that a staggered crack propagates diagonally between holes, increasing the fracture path length compared to a straight transverse crack.

The shear lag factor U accounts for the fact that not all elements of a cross-section participate equally in resisting tension at the connection. Per AISC 360 Table D3.1:

Block Shear (AISC 360 D5, J4.3)

Block shear is a gusset-plate-like failure mode where a block of material tears out around the bolt group. The failure surface consists of shear planes (parallel to the load, through the bolt holes) and a tension plane (perpendicular to the load, at the end of the bolt group). The nominal block shear strength is:

Rn = 0.60 x Fu x Anv + Ubs x Fu x Ant ≤ 0.60 x Fy x Agv + Ubs x Fu x Ant

where Anv is the net area in shear, Ant is the net area in tension, Agv is the gross area in shear, and Ubs = 1.0 when the tension stress is uniform (bolted connection with multiple tension bolts), or Ubs = 0.5 when the tension stress is non-uniform (single bolt in tension). The first term represents shear rupture along the hole lines; the second represents tension rupture at the end of the bolt group. The inequality limits the total to shear yield plus tension rupture, preventing double-counting.

For a double-angle tension member with 4 bolts through each leg, there are several potential block shear paths: (a) shear through both legs along the bolt line and tension across the end of both legs, (b) shear through one leg and tension through the gusset plate, and (c) shear through the gusset plate in the opposite direction. All paths must be checked.

Slenderness Limits (AISC 360 D1)

AISC 360 D1 recommends L/r ≤ 300 for tension members subject to stress reversal (wind uplift, seismic, vibration). For static tension members with no stress reversal (e.g., bottom chord of a roof truss under gravity only), there is no code-mandated slenderness limit, though practical considerations (sag and vibration during erection, handling, wind during construction) typically keep L/r under 400. For rods and bars, practical sag limits the length independently of r.

Worked Example -- Double-Angle Truss Bottom Chord

Problem: Check a double-angle tension bottom chord for a 100-ft span roof truss. Member: 2L6x6x1/2 (A36, Fy = 36 ksi, Fu = 58 ksi), single angle properties: A = 5.75 in^2, x_bar = 1.68 in. Connection: two lines of 7/8-inch bolts (four bolts per line) through each angle leg. Bolt spacing: 3 in pitch, 3-1/2 in gage. End distance: 2 in. Pu = 350 kips (factored). Member length = 50 ft (bottom chord between splices). LRFD design.

Step 1 -- Gross section yielding (D2). Ag = 2 x 5.75 = 11.50 in^2. phi_Pn = 0.90 x 36 x 11.50 = 373 kips. DCR_yield = 350/373 = 0.94. Passes, but close.

Step 2 -- Net section (D3). Bolt hole diameter: d_hole = 7/8 + 1/8 = 1.0 in. For each angle, 2 lines of holes at 3-1/2 in gage: An_per_angle = 5.75 - 2 x 1.0 x 0.50 (leg thickness) = 5.75 - 1.00 = 4.75 in^2. Total An = 2 x 4.75 = 9.50 in^2.

Shear lag: The angles are connected through one leg only (the leg bolted to gusset). For single angles with 4+ bolts, Table D3.1 Case 2 gives U = 0.80. But for double angles, the angles are on both sides of the gusset, improving the stress distribution. Use U = 1 - x_bar/L where L = distance from first to last bolt = 3 bolts x 3 in = 9 in. U = 1 - 1.68/9 = 1 - 0.187 = 0.813.

Ae = U x An = 0.813 x 9.50 = 7.72 in^2. phi_Pn = 0.75 x 58 x 7.72 = 336 kips. DCR_rupture = 350/336 = 1.04 -- FAILS (marginally).

Shear lag is reducing the effective area by nearly 19%, and combined with the bolt hole deduction of 17%, the net effective area is only 67% of the gross area. This 33% reduction is why rupture with U < 1.0 frequently governs angle tension members.

Step 3 -- Increase member or connection. Option A: Try 2L6x6x5/8 (A = 7.10 in^2 per angle, t = 5/8 in). Ag = 14.20 in^2. An = 14.20 - 4 x 1.0 x 0.625 = 14.20 - 2.50 = 11.70 in^2. U = 1 - 1.68/9 = 0.813 (same, x_bar changes slightly with thicker section, but approximately same). Ae = 0.813 x 11.70 = 9.51 in^2. phi_Pn_rupture = 0.75 x 58 x 9.51 = 414 kips. phi_Pn_yield = 0.90 x 36 x 14.20 = 460 kips. DCR_rupture = 350/414 = 0.85. Passes.

Option B: Keep the 1/2-inch angles but increase connection length (add 2 more bolts, L = 15 in, U = 1 - 1.68/15 = 0.888). Check hole count: 3 lines of holes = 3 x 1.0 x 0.50 = 1.50 in^2 deduction per angle. An = 5.75 - 1.50 = 4.25 in^2 per angle = 8.50 total. Ae = 0.888 x 8.50 = 7.55 in^2 (slightly less -- more holes hurt net area). U improvement of 9% is offset by additional hole area. Not effective here.

Conclusion: Use Option A: 2L6x6x5/8 with 8 bolts (4 per line, 2 lines per angle). Connection L = 9 in.

Step 4 -- Block shear (D5/J4.3). Shear path through bolt holes: L_shear = 3 x 3 (bolt spacing) + 2 (end distance) = 11 in per line. Two lines, two angles = 4 shear planes. Agv = 4 x 11 x 0.625 = 27.5 in^2. Anv = Agv - 4 x 3.5 holes x 1.0 x 0.625 = 27.5 - 4 x 2.188 = 27.5 - 8.75 = 18.75 in^2.

Tension path across end: width = bolt gage + bolt hole center to edge = 3.5 + 2 = 5.5 in. Two angles = 2 tension planes. Ant = 2 x (5.5 - 1.0) x 0.625 = 2 x 4.5 x 0.625 = 5.63 in^2. Agt (gross tension) = 2 x 5.5 x 0.625 = 6.88 in^2.

Rn = 0.60 x Fu x Anv + Ubs x Fu x Ant = 0.60 x 58 x 18.75 + 1.0 x 58 x 5.63 = 652 + 327 = 979 kips. Check upper bound: 0.60 x Fy x Agv + Ubs x Fu x Ant = 0.60 x 36 x 27.5 + 327 = 594 + 327 = 921 kips. Governing Rn = 921 kips. phi_Rn = 0.75 x 921 = 691 kips. DCR = 350/691 = 0.51. Passes.

Step 5 -- Slenderness check (D1). r_min for L6x6x5/8: rx = 1.84 in, ry_per_pair = 2.71 in (composite). r_z = 1.19 in (individual angle). L/r = 50 x 12 / 1.19 = 504 >> 300. FAILS slenderness limit for stress reversal.

If the bottom chord is subject to wind uplift reversal, the L/r = 504 exceeds 300. However, for a bottom chord with no stress reversal (gravity only), AISC D1 does not mandate a limit. Most designers would provide an intermediate sag rod or tie to reduce the unbraced length, or accept the higher slenderness with practical measures (erection bracing). For this example, assume the bottom chord is in static tension only (no stress reversal), so no slenderness limit applies.

Result: 2L6x6x5/8 (A36) bottom chord. Tension rupture governs at DCR = 0.85. Connection requires 8 bolts (7/8-inch diameter, 4 per line, 2 lines per angle) with 3 in pitch and 2 in end distance. Block shear is adequate.

Frequently Asked Questions

What is the difference between gross section yielding and net section rupture?

Gross section yielding (phi = 0.90) checks the full cross-section area against yield stress Fy. It is a ductile, serviceability-related limit state (excessive elongation). Net section rupture (phi = 0.75) checks the reduced section at bolt holes against ultimate tensile stress Fu. It is a brittle, strength limit state (fracture). The lower phi factor for rupture (0.75 vs 0.90) reflects the higher consequence of a sudden fracture. Proper tension member design requires that yielding governs over rupture (DCR_yield > DCR_rupture), ensuring ductile behavior where the member yields visibly before fracturing.

What is shear lag and how does the U factor work?

Shear lag occurs when tension forces are transferred through only some elements of a cross-section (e.g., only the connected leg of an angle, or only the flanges of a wide-flange). The force must "lag" through shear within the member to reach the unconnected elements, creating a non-uniform stress distribution. The U factor (≤ 1.0) converts the net area An to an effective net area Ae that accounts for this non-uniformity. For bolted connections, U = 1 - x_bar/L where x_bar is the connection eccentricity and L is the connection length. Longer connections (large L) reduce shear lag; the AISC minimum of L ≥ 4 bolts helps ensure U ≥ 0.75 for most connections.

What slenderness limits apply to tension members?

AISC 360-22 Section D1 recommends L/r ≤ 300 for tension members subject to stress reversal from wind, seismic, or vibration. The limit prevents excessive sag, vibration, and lateral deflection during handling and erection. For static tension members with no stress reversal, there is no code-mandated limit. However, AISC recommends L/r ≤ 300 as good practice for all tension members. Rods, cables, and flat bars in tension may have effective slenderness controlled by sag rather than L/r; for these, the 1/4-inch per foot of length sag limit is a practical criterion.

What is the s^2/(4g) rule for staggered bolt holes?

When bolt holes in adjacent lines are staggered, the net section may fail along a diagonal path rather than a straight transverse path. The diagonal path is longer, increasing the effective net width. AISC 360 Section B4.3b adds the term s^2/(4g) to the net width for each staggered pair, where s is the longitudinal stagger (pitch) and g is the transverse spacing (gage). For a 3-inch stagger at 3-1/2-inch gage: s^2/(4g) = 9/14 = 0.64 inches of recovered width per staggered pair. Strategic staggering can significantly improve net section capacity at minimal fabrication cost increase.

Which design standards cover tension member design and do they differ significantly?

AISC 360-22 Chapter D (US), AS 4100 Section 7 (Australia), EN 1993-1-1 Section 6.2.3 (Europe), and CSA S16 Section 13 (Canada) all cover tension member design. The fundamental limit states are the same in all codes: gross section yielding and net section rupture. The primary differences are in resistance factors (phi = 0.90/0.75 LRFD in US; gamma_M0 = 1.0 and gamma_M2 = 1.25 in Europe; phi = 0.90/0.80 in Australia; phi = 0.90/0.85 in Canada) and in shear lag treatment (AISC uses the U factor approach; EN 1993 uses a reduced net section based on connection category; AS 4100 uses a similar kt factor). Block shear treatment also varies: AISC uses the 0.60Fu x Anv + Ubs x Fu x Ant formulation, while EN 1993 uses a simpler effective net area approach per EN 1993-1-8.

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Disclaimer (Educational Use Only)

This page is provided for general technical information and educational use only. It does not constitute professional engineering advice. All structural designs must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) registered in the project jurisdiction. The site operator disclaims all liability for any loss or damage arising from the use of this page or the associated calculator tool. Results are preliminary -- not for construction.